Week 10: WS18 #1

1. Hill spheres: One outcome of planet formation is systems of satellites around planets. Now you may ask yourself, why do some planets have moons 100s of millions of kilometers away, while the Earth's moon is only 400,000km away? To answer this question, we need to think about how big of a region around a planet is dominated by the gravity of the planet, i.e. the region where the gravitational pull of the planet is more important than the gravitational pull of the central star (or another planet). 
a) Gravitational forces: Put a test mass somewhere between a star of mass M_s and a planet of mass m_p at a distance r_p from the star. Make a drawing clearly marking these characteristics for the gravitational force on the particle from the star and on the particle from the planet. At what distance r_H from the planet are the two forces balanced? This distance approximates the radius of the Hill sphere, which in the case of planet formation is the sphere of disk material which a planet can accrete from. 

We know that gravitational force is given by \(F_G=\frac{GMm}{r^2}\), so we can just set the gravitational force on the particle from the star and from the planet equal and solve for r_H. 

\(\frac{GM_sm}{(r_p-r_H)^2}=\frac{Gm_pm}{r_H^2}\) 

\(\frac{M_s}{m_p}=(\frac{r_p-r_H}{r_H})^2=(\frac{r_p}{r_H}-1)^2\) 

\(\sqrt{\frac{M_s}{m_p}}=\frac{r_p}{r_H}-1\)

\(r_H=\frac{r_p}{1+\sqrt{\frac{M_s}{m_p}}}\)

b) Planetary Hill radii: Calculate the Hill radii for Earth, Jupiter, and Neptune. How do they compare with the separation between the planets and their most distant moons?

We can use the table above to calculate the Hill radii for Earth, Jupiter, and Neptune. Since the masses are given to us in terms of Earth masses, it's helpful to know that the Sun is 333,000 times as massive as Earth. 

\(r_{H(Earth)}=\frac{1AU}{1+\sqrt{\frac{3.33\times10^5M_\oplus}{1M_\oplus}}}=0.0017AU=2.59\times10^{5}km\)

\(r_{H(Jupiter)}=\frac{5AU}{1+\sqrt{\frac{3.33\times10^5M_\oplus}{320M_\oplus}}}=0.150AU=2.25\times10^{7}km\)

\(r_{H(Neptune)}=\frac{30AU}{1+\sqrt{\frac{3.33\times10^5M_\oplus}{17M_\oplus}}}=0.213AU=3.18\times10^{7}km\)

These numbers are fairly consistent with the observed moon distances for Earth, Jupiter, and Neptune.